College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter 1 - Review Exercises - Page 163: 43


$x=\left\{ 2-\sqrt{6},2+\sqrt{6} \right\}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ (2x+1)(x-4)=x ,$ express first in the form $ax^2+bx+c=0.$ Then use the Quadratic Formula to solve for $x.$ $\bf{\text{Solution Details:}}$ Using the FOIL Method which is given by $(a+b)(c+d)=ac+ad+bc+bd,$ the expression above is equivalent to\begin{array}{l}\require{cancel} 2x(x)+2x(-4)+1(x)+1(-4)=x \\\\ 2x^2-8x+x-4=x .\end{array} In the form $ax^2+bx+c=0,$ the equation above is equivalent to \begin{array}{l}\require{cancel} 2x^2+(-8x+x-x)-4=0 \\\\ 2x^2-8x-4=0 .\end{array} In the equation above, $a= 2 ,$ $b= -8 ,$ and $c= -4 .$ Using the Quadratic Formula which is given by $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a},$ then \begin{array}{l}\require{cancel} x=\dfrac{-(-8)\pm\sqrt{(-8)^2-4(2)(-4)}}{2(2)} \\\\ x=\dfrac{8\pm\sqrt{64+32}}{4} \\\\ x=\dfrac{8\pm\sqrt{96}}{4} \\\\ x=\dfrac{8\pm\sqrt{16\cdot6}}{4} \\\\ x=\dfrac{8\pm\sqrt{(4)^2\cdot6}}{4} \\\\ x=\dfrac{8\pm4\sqrt{6}}{4} \\\\ x=\dfrac{\cancel4^2\pm\cancel4^1\sqrt{6}}{\cancel4^1} \\\\ x=2\pm\sqrt{6} .\end{array} The solutions are $ x=\left\{ 2-\sqrt{6},2+\sqrt{6} \right\} .$
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